$\begingroup$Note that $X$ is $T_1$ iff $Y$ is $T_1$: If $X$ is $T_1$ and $f(x),f(y)$ are points of $Y$, then by local injectivity, $x$ has a neighborhood $U$ containing at most one point from the fiber of $y$, and we can shrink $U$ such that it does not contain that point any more. Now $f(U)$ is a neighborhood of $f(x)$ disjoint from $f(y)$. For the other direction, if $x$ and $y$ are points in $X$, and if $f(x)\ne f(y)$, then a neighborhood of $f(x)$ disjoint from $f(y)$ pulls back to such a neighborhood of $x$. And if $f(x)=f(y)$, then local injectivity gives a neighborhood of $x$ not containing $y$.$\endgroup$
– Stefan HamckeJun 30 '15 at 16:33

1

$\begingroup$One sufficient condition for the existence of disjoint nbh of the points in $F=f^{-1}(y)$ would be $X$ being $T_3$ and fibers being countable. In that case, $Y$ is $T_1$, so $y$ is closed, implying that any subset of $F$ is closed in $X$. Assume by induction we have found pairwise disjoint neighborhoods $U_1,\dots, U_n, U^{n+1}$ of the first $n$ points in $F=\{x_1,x_2,\dots\}$ and of $\{x_{n+1},\dots\}$, respectively. Now choose a nbh $U_{n+1}$ of $x_{n+1}$ within $U^{n+1}$ and disjoint from a neighborhood $U^{n+2}$ of $\{x_{n+2},\dots\}$. This way, we get pairwise disjoint neighborhoods.$\endgroup$
– Stefan HamckeJun 30 '15 at 16:35

$\begingroup$I don't know if this suits you since the conditions are not particularly weak.$\endgroup$
– Stefan HamckeJun 30 '15 at 16:36

$\begingroup$Do you have a counterexample where it fails?$\endgroup$
– HeikoJun 30 '15 at 21:05

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$\begingroup$Only one which is not Hausdorff: Let $X=\{a,b,c\}$ with the particular point topology of $b$, and let $Y=\{d,e\}$ with the particular point topology of $e$ (this is also called the Sierpinski space). The map $f:X\to Y$ sending $a$ and $c$ to $d$, and $b\mapsto e$, is a local homeomorphism, but $a$ and $c$ cannot be separated by neighborhoods.$\endgroup$
– Stefan HamckeJun 30 '15 at 23:41